I have yet to see any problem, however complicated, which, when looked at in the right way did not become still more complicated, Paul Anderson.

Ned, I would love to stand here and talk with you—but I’m not going to, Phil Connors (Bill Murray), Groundhog Day

I don’t have to take this abuse from you; I’ve got hundreds of people dying to abuse me, Dr. Peter Venkman (Bill Murray), Ghostbusters

Recall

Definition. A differential equation is an equation that involves one or more dependent variables, their derivatives with respect to one or more independent variables, and the independent variables themselves, e.g., $\frac{dy}{dx} = 3x +5y, y’ + y = 4xcos(2x), \frac{dy}{dx} = x^2y+y, etc.$

It involves (e.g., $\frac{dy}{dx} = 3x +5y$):

Dependent variables: Variables that depend on one or more other variables (y).
Independent variables: Variables upon which the dependent variables depend (x).
Derivatives: Rates at which the dependent variables change with respect to the independent variables, $\frac{dy}{dx}$

First-Order Linear Ordinary Differential Equations (ODEs)

Definition. A first-order linear ordinary differential equation is an ordinary differential equation (ODE) involving an unknown function y(x), its first derivative y′, and functions of the independent variable x, which can be written in the general form:: a(x)y' + b(x)y = c(x) where:

y = y(x) is the unknown function of the independent variable x.
y′ = $\frac{dy}{dx}$ is the derivative of y with respect to x.
a(x), b(x), and c(x) are known functions of x.

These equations are termed “linear” because the unknown function y and its derivative y’ appear to the first power and are not multiplied together or composed in any nonlinear way.

If the function c(x)=0 for all x in the interval of interest, the equation simplifies to: a(x)y’ + b(x)y = 0. Such an equation is called a homogeneous linear differential equation.

The Existence and Uniqueness Theorem provides crucial insight into the behavior of solutions to first-order differential equations ODEs. It states that if:

The function f(x, y) (the right-hand side of the ODE) in the equation y’ = f(x, y) is continuous in some neighborhood around a point (x₀, y₀) and
Its partial derivative with respect to y, denoted as $\frac{∂f}{∂y}$, is also continuous near (x₀, y₀).

Then, the differential equation y' = f(x, y) has a unique solution to the initial value problem through the point (x₀, y₀), meaning that it satisfies the initial condition y(x₀) = y₀.

This theorem ensures that under these conditions, the solution exists and is unique near x = x₀.

First-order Autonomous ODE’s

Definition. A first order autonomous ordinary differential equation (ODE) is a differential equation in which the rate of change of a dependent variable depends solely on y itself, and not explicitly on the independent variable, typically denoted by t. In mathematical terms, it has the general form: $\frac{dy}{dt}=f(y)$.

Here, f(y) is a function that depends only on y, and there is no explicit dependence on the independent variable t. This absence of explicit dependence on t is what characterizes the equation as autonomous.

Characteristics of Autonomous ODEs

Because the right-hand side of the equation depends only on y, autonomous ODEs exhibit unique properties that distinguish them from non-autonomous equations:

Time Translation Invariance: The solutions to autonomous ODEs are invariant under shifts in the independent variable t. This means that if y(t) is a solution, then y(t −t₀) is also a solution for any constant t₀. This property arises because the equation does not depend explicitly on t, so shifting t by a constant does not affect the form of the solution.
Separable Variables: Autonomous ODEs are inherently separable, allowing us to rewrite the equation in the form: $\int \frac{1}{f(y)}dy = \int dt + C$. This separability facilitates solving the ODE by integrating with respect to y and t.

Challenges in Solving Autonomous ODEs

While autonomous ODEs are separable, the integration may not always be straightforward due to:

Complex Integrals: The function f(y) might lead to integrals that are difficult or impossible to evaluate in terms of elementary functions.
Implicit Solutions: Sometimes, the solution can only be expressed implicitly, meaning that y is not isolated on one side of the equation. This can make it challenging to interpret the solution or compute specific values for y given t.

In such situations, qualitative analysis becomes a valuable tool to gain insights into the behavior of the solutions without solving the equation analytically.

Qualitative Analysis of Autonomous ODEs

In qualitative analysis, we study the direction field of the ODE. This field helps visualize the behavior of solutions without explicitly solving the equation.

A direction field (or slope field) is a graphical representation of the ODE that shows the slope $\frac{dy}{dt}$ at various points in the (t, y) plane. By plotting small line segments with appropriate slopes at several points, we can visualize how solutions to the ODE behave.

Key Concepts in Qualitative Analysis

Horizontal Isoclines. In an autonomous ODE, since $\frac{dy}{dt} = f(y)$ the slope at any point depends only on the value of y. Therefore, along any horizontal line y = y₀, the slope $\frac{dy}{dt}$ is constant. Specifically, the slope along this line is: $\frac{dy}{dt} = f(y_0)$. Since f(y₀) is just a constant (with no t-dependence), the slopes along the line y = y₀ are the same, making the line an isocline —a line where the slopes are constant (Refer to Figure E for a visual representation and aid in understanding it). This means that solutions to the ODE, called integral curves, will behave similarly along these horizontal lines.
Integral curves are invariant under horizontal translation: Shifting the solution curves horizontally (along the t-axis) does not change their shape because f(y) does not depend on t. The solutions will look the same at different times, reflecting the time translation invariance of autonomous ODEs.

First Order Autonomous ODEs

Behavior Based on f(y). When f(y) > 0, the solution (the slope $\frac{dy}{dt}$ is positive) y(t) is increasing at that point. When f(y) < 0, the solution (the slope $\frac{dy}{dt}$ is negative) y(t) is decreasing at that point.
A critical point or equilibrium solution occurs where f(y) = 0. At these points, the rate of change $\frac{dy}{dt} = 0$,meaning that the solution remains constant over time.

If y₀ is a critical point, then y(t) = y₀ is a solution to the differential equation because: $\frac{dy}{dt}|_{y=y_0}=\frac{dy_0}{dt}=f(y_0)=0$. The left hand size is zero because the derivative of a constant is zero, and the right hand side is zero because f(y₀) = 0 by definition of a critical point.

The Existence and Uniqueness Theorem implies that solutions to the ODE cannot cross each other in the (t, y) plane. Specifically, solutions starting between two critical points y = y₀ and y = y₁cannot cross these horizontal lines. This ensures that solutions maintain consistent behavior and do not overlap (Refer to Figure 1 for a visual representation and aid in understanding it).

First Order Autonomous

Steps for Qualitative Analysis of Autonomous ODEs

To analyze an autonomous ODE, follow these steps:

Find the critical points. Solve f(y) = 0 to identify where the solutions remain constant (i.e., equilibrium solutions).
Analyze the sign of f(y) and draw the Phase Line. Sketch or plot f(y) versus y to visualize where f(y) is positive, negative, or zero.

When f(y) > 0, the slope $\frac{dy}{dt}$ is positive, meaning that the solution y(t) is increasing. When f(y) < 0, the slope $\frac{dy}{dt}$ is negative, meaning that the solution y(t) is decreasing.
Determine Stability of Critical Points: Stable Equilibrium (Solutions near the critical point move towards it as t increases. The point is attracting), Unstable Equilibrium (Solutions near the critical point move away from it as t increases. The point is repelling), or Semi-Stable Equilibrium (The critical point is stable on one side and unstable on the other).
Interpret the Results. Discuss the long-term behavior of solutions. Determine whether solutions approach a critical point, diverge to infinity, or exhibit other behaviors based on initial conditions.
Consider initial conditions. Analyze how different initial values y(0) affect the solution’s trajectory and its tendency to move towards or away from critical points.

Example $\frac{dy}{dt} = 1-y^2$

Autonomous Nature. This equation is a first order linear autonomous ODE because it relates the derivate $\frac{dy}{dt}$ to the function y(t) itself and does not explicitly depend on the independent variable t.

Finding Critical Points. Equilibrium points occurs where $\frac{dy}{dt} = 0$, $\frac{dy}{dt} = 0 ↭ 1-y^2 = 0↭ (1-y)(1+y) = 0 ↭ y = ±1$.

Analyzing the Sign of f(y). To understand this differential equation, we analyze the sign of $\frac{dy}{dt}$ for values of y above and below the critical point.

For y < -1: $\frac{dy}{dt} = 1 -y^2 < 0$. So, y decreases.
For -1 < y < 1: $\frac{dy}{dt} = 1 -y^2 > 0$. So, y increases.
For y > 1: $\frac{dy}{dt} = 1 -y^2 < 0$. So, y decreases.

Drawing the Phase Line. Arrows indicating the direction of y(t) point upwards (towards y = 1, solutions increase) from -1 < y < 1, indicating attraction, so y = 1 is a stable equilibrium point.

Slopes Between y = −1 and y = 1: (1) Start at Zero at y = −1 ($\frac{dy}{dt} = 0$). (2) Between y = -1 and y = 0, slopes increase from 0 to a maximum of 1 ($\frac{dy}{dt}=1-y²\bigg|_{y=0}=1-0 = 1$) at y = 0. (3) Between y = 0 and y = 1, slopes decrease back from its maximum 1 to zero (y = 1). Solutions increase fastest near y = 0. The rate of increase slows down as y approaches y = 1.

Below y = -1 (y < -1) and above y = 1 (y > 1), slopes are negative, solutions y(t) decreases as t increases, arrows point downwards, that is, away from y = -1 and towards y = 1, respectively, hence y = -1 is a unstable equilibrium point.

Plots of y(t) for different initial conditions illustrate the convergence towards y = 1 for ∣y₀| < 1 and y₀ > 1, and divergence for y₀ < -1 (Refer to Figure i for a visual representation and aid in understanding it)

First Order Autonomous ODEs

Solving the Differential Equation

This is an autonomous, separable first-order ordinary differential equation (ODE): $\frac{dy}{1-y^2} = dt$.

We perform partial fraction decomposition: $\frac{dy}{1-y^2} = \frac{1}{2}(\frac{1}{1+y}+\frac{1}{1-y})$

$\frac{1}{1.y^2} = \frac{1}{(1-y)(1+y)} = \frac{A}{1-y} + \frac{B}{\frac{1}{1+y}}↭ 1 = A(1+y)+B(1-y) = A(1)+A(y)+B(1)−B(y)=(A+B)+(Ay−By)↭[\text{Group like terms}] A + B = 1, A - B = 0$. From (ii), A = B. From (1), 2A = 1, $A = \frac{1}{2}, B = A = \frac{1}{2}$.

Integrating both sides: $\int \frac{1}{2}(\frac{1}{1+y}+\frac{1}{1-y})dy = \int dt$.

This yields: $\frac{1}{2}ln|1+y| -\frac{1}{2}ln|1-y| = t + C ↭ ln|1+y| -ln|1-y| = 2(t + C) ⇒[\text{Combining logarithms:}] ln(\frac{1+y}{1-y}) = 2(t+C) ⇒[\text{Exponentiating both sides of the equation}] \frac{1+y}{1-y} = e^{2(t+C)} = Ae^{2t}$ where A = e^2C.

$\frac{1+y}{1-y} = Ae^{2t} ↭ 1+y = Ae^{2t}(1-y)↭ 1 + y = Ae^{2t} -Ae^{2t}y ⇒[\text{Collect like terms}] y(1+Ae^{2t}) = Ae^{2t} -1⇒[\text{Solve for y}] y = \frac{Ae^{2t} -1}{Ae^{2t} +1}$. This is the explicit solution for y(t).

The constant A can be found using an initial condition y(0) = y₀, $y_0 = \frac{A-1}{A+1}↭ (A+1)y_0 = A -1 ↭ Ay_0 + y_0 = A - 1 ↭ Ay_0 -A = -y_0 -1 ↭ A(y_0 -1) = -y_0 -1 ↭ A = \frac{-y_0-1}{y_0-1}↭ A = \frac{y_0+1}{1-y_0}$

The solution to the differential equation is: $\frac{Ae^{2t} -1}{Ae^{2t} +1}$ where $A = \frac{y_0+1}{1-y_0}$

Analyzing the Limits, $\lim_{t \to ∞} y(t) = \lim_{t \to ∞} \frac{Ae^{2t} -1}{Ae^{2t} +1} = 1$ (the terms -1 and 1 become negligible compared to Ae^2t).

Regardless of the value of A (provided A > 0), y(t) approaches 1 as t → ∞, confirming that y = 1 is a stable equilibrium. Besides, $\lim_{t \to -∞} y(t) = \lim_{t \to -∞} \frac{Ae^{2t} -1}{Ae^{2t} +1} = \frac{0-1}{0+1} = -1$, which aligns with the instability of y = −1 in forward time (y = -1 is an unstable equilibrium) but shows attraction in backward time.

Example: Bank Account with Embezzlement

Let’s explore how to model and analyze the behavior of a bank account that is simultaneously accruing interest and experiencing consistent withdrawals due to embezzlement. This scenario helps illustrate how differential equations can be used to understand real-world financial situations.

Let’s consider the following scenario where:

y(t) is the amount of money in a bank account at time t.
r is the continuous annual interest rate on the account.
w is the constant rate at which money is being embezzled from the account (the amount of money being stolen per unit time, e.g., dollars per day, euros per hour, etc.).
Suppose a bank teller is illicitly withdrawing funds from a customer’s bank account without their knowledge or voluntary consent. The customer continues to accrue interest on their account balance at a continuous rate r, but the ongoing embezzlement at rate w affects the overall growth of the account. We aim to model the evolution of the account balance over time and understand under what conditions the balance will increase or decrease.

Bank Account with Embezzlement

Mathematical Modeling

The rate of change of the account balance y(t) over time can be expressed as the difference between the interest earned and the amount being embezzled: $\frac{dy}{dt} = ry -w$ where:

ry: Represents the rate at which the account balance increases due to continuous interest. The growth is proportional to the current balance y(t).
−w: Represents the constant rate at which money is being embezzled from the account, reducing the balance.

Analysis of the ODE

Autonomous Nature: This equation is a first-order linear autonomous ODE because it relates the derivative $\frac{dy}{dt}$ to the function y(t) itself and does not explicitly depend on the independent variable t.
An autonomous ODE is one where the rate of change of the dependent variable depends solely on the dependent variable itself and not explicitly on the independent variable.
Finding Critical Points (Equilibrium Solutions): Critical points occur when the rate of change of the dependent variable is zero: $\frac{dy}{dt} = 0 ⇒ ry -w = 0 ↭[\text{Solving for y:}] y = \frac{w}{r}$ (Refer to Figure 2 for a visual representation and aid in understanding it). Hence, $y = \frac{w}{r}$ is the equilibrium solution or critical point.
At this balance, the amount of interest earned exactly offsets the amount being embezzled, resulting in a net zero change in the account balance over time.

Arrows indicating the direction of y(t). Upwards for y > w/r, and downwards for y < w/r
Behaviour of the solution. To understand how the account balance changes over time, we analyze the sign of $\frac{dy}{dt}$ for values of y above and below the critical point.

When $y > \frac{w}{r}$, the function ry -w is positive, meaning $\frac{dy}{dt} > 0$. The solution is increasing, which means the bank account balance grows over time.
If the bank account has more than w/r (the critical point), the account balance increases over time because the interest earned per unit time exceeds the embezzlement rate. The customer may not notice the embezzlement immediately due to the increasing balance.
If $y < \frac{w}{r}$, the function ry -w is negative, meaning $\frac{dy}{dt} < 0$. The solution is decreasing, meaning that the bank account balance is shrinking over time.
If the bank account has less than w/r (the critical point), the account balance diminishes because the losses due to embezzlement outweigh the interest gains. The account may eventually deplete, potentially leading to financial issues.

Solving the Differential Equation Explicitly

Separate Variables: $\frac{dy}{dt} = ry-w ↭ \frac{dy}{ry-w} = dt$
Integrate Both Sides: $\int \frac{dy}{ry-w} = \int dt ↭[u = ry -w⇒ du = rdy] \frac{1}{r}\int \frac{du}{u} = t + C ↭ \frac{1}{r}ln|u| = t + C ↭ \frac{1}{r}ln|ry-w| = t + C$
Solve for y. Multiply both sides by r: ln|ry -w| = rt + rC. Exponentiate both sides to eliminate the natural logarithm: $|ry -w| = e^{rt}e^{rC}↭ ry -w = ±e^{rC}e^{rt} ↭ ry -w = ±Ae^{rt}$ where $A = e^{rC}$ is a constant ↭ $ry = w ±Ae^{rt} ↭ y = \frac{w}{r} ±\frac{A}{r}e^{rt} ↭ y = \frac{w}{r} + Ce^{rt}$ where $C = ±\frac{A}{r}$
Apply initial conditions, y(0) = y₀, $y_0 = \frac{w}{r}+ Ce^{r·0}↭ y_0 = \frac{w}{r}+ C ⇒ C = y_0 -\frac{w}{r}$
Final Explicit Solution: $y(t) = \frac{w}{r}+(y_0-\frac{w}{r})e^{rt}$

Interpretation:

If $y_0 > \frac{w}{r} ⇒ C > 0$. As t → ∞, $e^{rt}→ ∞$, y(t) → ∞. The account balance grows exponentially over time (without bound). The interest earned not only compensates for the embezzlement but also contributes to substantial growth of the account. The customer may not notice the embezzlement immediately due to the increasing balance.
If $y_0 = \frac{w}{r} ⇒ C = 0 ⇒ y(t) = \frac{w}{r}$. The account balance remains steady. The customer might not notice any changes in the account balance, masking the embezzlement.
If $y_0 < \frac{w}{r} ⇒ C < 0$. As t → ∞, $e^{rt}→ ∞$, y(t) → -∞. However, in practical terms, the account cannot have an infinite negative balance. Practically, account balance will decrease and may reach zero, potentially leading to overdrafts or financial penalties. The customer may face financial issues due to the decreasing balance and potential overdraft fees.

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This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License and is based on MIT OpenCourseWare [18.01 Single Variable Calculus, Fall 2007].

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